Applications of Boyle's law
a)As we know from Boyle's law, PV = constant. With the increase in altitude, the pressure of the gas decreases which causes the decrease in the density of gas decreases. So, mountain climbers may face a problem called altitude sickness. To recover this problem, they carry oxygen gas cylinder with them.
b)From Boyle's law, pressure and volume have an inverse relation. With the increase in altitude, the pressure decreases which in turn increases the volume of the gas. Therefore, the size of air balloon goes on increasing as it ascends the height.
Charle’s law
Charle’s law states that “ The volume of a given mass of given mass increase or decrease by (\(\frac{1}{273}\) )^{th} of its volume at 0^{0}C for every ^{0}C rise or fall in temperature”.
Let V_{o} be the volume of the gas at 0^{0}C,
Volume of the gas at 1^{0}C
= V_{o }+ V_{o} \(\frac{1}{273}\)
= \( \frac {V_o ( 1 + 273)}{273}\) at 1^{0}C
Volume of the gas at 3^{0}C
= V_{o }+ V_{o} \(\frac{3}{273}\)
= \( \frac {V_o (3 + 273)}{273}\) at 3^{0}C
Volume of the gas at t^{0}C
= V_{o }+ V_{o} \(\frac{t} {273}\)
= \( \frac {V_o (t + 273)}{273}\) at t^{0}C
Let 'V' be the final temperature of the gas at 't^{0}C',
Then, V = \( \frac {V_o (t + 273)}{273}\) -------------(i)
Equation (i) is the mathematical form of Charle's law.
Verification of Charle's law
Rearranging equation (i)
V = V_{0} + V_{0 }\( \frac{t}{273}\)
or, V = \( \frac{V_o}{273}\) t + V_{0} -(ii)
Equation (ii) is in the form of y = mx + c. So, a graph of 'V' vs 't' gives a straight line with slope \( \frac{V_o}{273}\) and y-intercept 'V_{0}'
Alternative form of Charle's law
Charle’s law states that “ The volume of a given mass of given mass increase or decrease by (\(\frac{1}{273}\) )^{th} of its volume at 0^{0}C for every ^{0}C rise or fall in temperature”.
Mathematically,
V = \( \frac {V_o (t + 273)}{273}\) -------------(i)
Let, V_{1} and V_{2} be the volume of given mass of gas at t_{1}^{0}C and t_{2}^{0}C respectively. Then, from equation (i)
V_{1} =\( \frac {V_o (t_1 + 273)}{273}\) ----(ii)
V_{2 }=\( \frac {V_o (t_2 + 273)}{273}\) ----(iii)
Dividing euqation (ii) by (iii)
\( \frac {V_1}{V_2}\) = \( \frac {273 + t_1}{273 + t_2}\)
Since, 273 + t_{1} = T_{1} and 273 + t_{2} = T_{2}
\( \frac {V_1}{V_2}\) =\( \frac {T_1}{T_2}\)
∴ V∝ T at constant pressure
From above expression, Charles law can also be regarded as " pressure remaining constant, volume of a given mass of gas is directly proportional to the absolute temperature"
Verification,
V∝ T
or, V = kT
This equation is the form of y = mx. So, a graph of 'V' vs "T' gives a straight line passing through origin as shown in the figure below.
Absolute zero ( Kelvin zero) and Absolute scale of temperature
From Charle's law
V =\( \frac {V_o (t + 273)}{273}\)
when, t = -273^{0}C
V = V_{o} - V_{0} = 0
At -273^{0}C, the volume of the gas becomes theoretically 0. All the gases become solid or liquid before this temperature is reached. This hypothetical temperature is known as absolute zero. It is denoted by OA, OK, etc.
The temperature scale based on the absolute zero as its starting point is known as Kelvin scale of temperature or absolute scale of temperature.
Relation between density and temperature of gas
From definition of density
Density (D) = \( \frac{Mass(M)}{Volume(V)}\)
From Charles law,
V = kT, where '' is a proportionality constant
or, \( \frac {M}{D}\) = kT
or, \( \frac {M}{k}\) = DT
For a given mass of gas, the value of 'm' is constant. So, \( \frac{M}{k}\) also gives another constant value
Then, TD = constant
Hence, T_{1}D_{1} = T_{2}D_{2}............T_{n}D_{n} at constant pressure
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